Problem: If $x \veebar y = (3-x)(y)$ and $x \diamond y = x^{2}-4y^{2}$, find $(-4 \diamond 1) \veebar -1$.
Solution: First, find $-4 \diamond 1$ $ -4 \diamond 1 = (-4)^{2}-4(1^{2})$ $ \hphantom{-4 \diamond 1} = 12$ Now, find $12 \veebar -1$ $ 12 \veebar -1 = (3-12)(-1)$ $ \hphantom{12 \veebar -1} = 9$.